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igor_vitrenko [27]
2 years ago
13

Help with 3 and check the other ones please :(

Physics
2 answers:
devlian [24]2 years ago
7 0
I don’t know what your talking about. ?!?!?!?
alekssr [168]2 years ago
7 0
Um there’s no pictures on here for me to help you with if you put a picture maybe I’ll be able to help you
You might be interested in
What will happen to plant height if the amount of available light is reduced due to global dimming?
SpyIntel [72]
The plants would wilt and decrease in height due to the lack of sunlight. I hope I helped!
7 0
3 years ago
A woman on a bicycle traveling at 10 m/s on a horizontal road stops pedaling as she starts up a hill inclined at 4.0º to the hor
IrinaK [193]
The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.

m mass
v velocity
h height
g = 9.81m/s²

The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)

6 0
3 years ago
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0
3 years ago
А A pool of water of refractive index
babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent  depth when viewed vertically through  air.​ The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm

So, the apparent depth is 45 cm.

3 0
2 years ago
A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3
marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

g(H-h)=\frac{1}{2}v^2

\Rightarrow H=\frac{v^2}{2g}+h

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

H=\frac{13.9^2}{2\times3}+3

H=35.20 m

4 0
3 years ago
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